The ShiftEquivalent command determines whether the two polynomials $f,g$ are shift equivalent w.r.t. the variable x, that is, whether there is an $h$ independent of x satisfying $\mathrm{lc}\left(g\right)f\left(x+h\right)=\mathrm{lc}\left(f\right)g\left(x\right)$, where $\mathrm{lc}$ denotes the leading coefficient with respect to x. It returns $h$, if it exists, and otherwise FAIL.

If the optional argument T is specified, then ShiftEquivalent returns FAIL even if $h$ exists but is not of type T. This is more efficient than first calling ShiftEquivalent without the optional argument and then checking whether the return value is of type T.

It is assumed that both input polynomials are collected w.r.t. the variable x.

If $f,g$ are nonconstant w.r.t. x, then $h$ is uniquely determined. If both are nonzero and constant w.r.t. $x$, or if both are zero, then the return value is 0.

$\mathrm{with}\left(\mathrm{PolynomialTools}\right)\:$

$\mathrm{ShiftEquivalent}\left(x^2+x+1\,x^2-x+1\,x\right)$

$\mathrm{-1}$

$\mathrm{Translate}\left(x^2+x+1\,x\,\right)$

$x^2-x+1$

$\mathrm{ShiftEquivalent}\left(x^2+1\,x^2-x+1\,x\right)$

$\mathrm{FAIL}$

$\mathrm{ShiftEquivalent}\left(2x-1\,x+\frac{1}{2}\,x\right)$

$1$

$\mathrm{Translate}\left(2x-1\,x\,\right)$

$1+2x$

$\mathrm{ShiftEquivalent}\left(2x-1\,x\,x\right)$

$\frac{1}{2}$

$\mathrm{ShiftEquivalent}\left(2x-1\,x\,x\,'\mathrm{integer}'\right)$

$\mathrm{FAIL}$

$\mathrm{ShiftEquivalent}\left(x\,x+n\,x\right)$

$n$